Question

Pls give solution with steps​

Answer 1

$\large\underline{\sf{Given- }}$

$\rm :\longmapsto\: \alpha , { \alpha}^{2}, { \alpha }^{3}, { \alpha }^{4} \: are \: roots \: of \: {x}^{5} - 1 = 0$

$\large\underline{\sf{To\:prove - }}$

$\rm :\longmapsto\:(1 - \alpha )(1 - { \alpha }^{2})(1 - { \alpha }^{3})(1 - { \alpha }^{4}) = 5$

$\begin{gathered}\Large{\sf{{\underline{Formula \: Used - }}}}  \end{gathered}$

$\underbrace{\boxed{ \tt{ \displaystyle\lim_{x \to a}\: \frac{ {x}^{n} - {a}^{n} }{x - a} = {na}^{n - 1} }}}$

$\large\underline{\sf{Solution-}}$

Given equation is

$\rm :\longmapsto\: {x}^{5} - 1 = 0$

By hit and trial, if we substitute x = 1, it satisfy the given equation.

So, x = 1 is root of equation.

And also, it is given that

$\rm :\longmapsto\: \alpha , { \alpha}^{2}, { \alpha }^{3}, { \alpha }^{4} \: are \: roots \: of \: {x}^{5} - 1 = 0$

Since, highest degree of given Equation is 5. So, it has atmost 5 real roots.

So, it implies,

$\rm :\longmapsto\: 1,\alpha , { \alpha}^{2}, { \alpha }^{3}, { \alpha }^{4} \: are \: roots \: of \: {x}^{5} - 1 = 0$

So,

$\rm :\longmapsto\: {x}^{5} - 1 = (x - 1)(x - \alpha)(x - { \alpha }^{2})(x - { \alpha }^{3})(x - { \alpha }^{4})$

So, it can be rewritten as

$\rm :\longmapsto\: \dfrac{ {x}^{5} - 1}{x - 1} = (x - \alpha)(x - { \alpha }^{2})(x - { \alpha }^{3})(x - { \alpha }^{4})$

So,

$\rm :\longmapsto\: \displaystyle\lim_{x \to 1}\dfrac{ {x}^{5} - 1}{x - 1} =\displaystyle\lim_{x \to 1} (x - \alpha)(x - { \alpha }^{2})(x - { \alpha }^{3})(x - { \alpha }^{4})$

$\rm :\longmapsto\: 5 {(1)}^{5 - 1} = (1 - \alpha)(1 - { \alpha }^{2})(1 - { \alpha }^{3})(1 - { \alpha }^{4})$

$\rm :\longmapsto\: 5 {(1)}^{4} = (1 - \alpha)(1 - { \alpha }^{2})(1 - { \alpha }^{3})(1 - { \alpha }^{4})$

$\rm :\longmapsto\: (1 - \alpha)(1 - { \alpha }^{2})(1 - { \alpha }^{3})(1 - { \alpha }^{4}) = 5$

Hence, Proved

Additional Information :-

$\underbrace{\boxed{ \tt{\displaystyle\lim_{x \to 0} \: \frac{sinx}{x} = 1}}}$

$\underbrace{\boxed{ \tt{\displaystyle\lim_{x \to 0} \: \frac{tanx}{x} = 1}}}$

$\underbrace{\boxed{ \tt{\displaystyle\lim_{x \to 0} \: \frac{ {e}^{x} - 1}{x} = 1}}}$

$\underbrace{\boxed{ \tt{\displaystyle\lim_{x \to 0} \: \frac{ log(1 + x) }{x} = 1}}}$

$\underbrace{\boxed{ \tt{\displaystyle\lim_{x \to 0} \: \frac{ {a}^{x} - 1}{x} = loga}}}$